A Young’s double‑slit experiment is performed with a slit separation of $0.20,\text{mm}$, a light of wavelength $520,\text{nm}$, and the screen is $1.20,\text{m}$ away. What is the fringe spacing β between consecutive bright fringes?

Step-by-step solution

1. Use the formula for fringe spacing in Young’s experiment: $\beta =\frac{\lambda L}{d}$.
2. Convert the slit separation $d=0.20,\text{mm}=2.0\times 10^{-4},\text{m}$, wavelength $\lambda =520,\text{nm}=5.20\times 10^{-7},\text{m}$, and distance $L=1.20,\text{m}$ into SI units.
3. Substitute into $\beta =\frac{5.20\times 10^{-7},\text{m}\times 1.20,\text{m}}{2.0\times 10^{-4},\text{m}}$.
4. Calculate numerator: $5.20\times 10^{-7}\times 1.20=6.24\times 10^{-7},{\text{m}}^2$.
5. Divide by $2.0\times 10^{-4},\text{m}$ to get $\beta =3.12\times 10^{-3},\text{m}$.
6. Convert to millimetres: $3.12\times 10^{-3},\text{m}=3.12,\text{mm}$.

Concept tested: Young’s double‑slit fringe spacing

Published 21 September 2026 · How this solution was prepared and checked